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Estimating activity of a radiocative sample

  1. May 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A sample of processed waste from a nuclear reactor contains 1.0*10^24 plutonium-239 atoms. The half life of plutonium-239 is 2.41*10^4 years.
    a) How many plutonium-239 atoms will decay in the next 2.41*10^4 years?
    b) Estimate the activity of the original sample.

    2. Relevant equations
    I guess that A=Ao(1/2)^t/(t*o.5)nwould be relevant, but really I don't see how. A= the decaying quantity remaining, which I know anyway, and nothing else that I could rearrange the equasion for will tell me the right answer.


    3. The attempt at a solution
    I managed to do a) alright, but I can't figure out b).
    I can't really attempt the solution because I have no idea how. I know that the activity will be divided by 2 after each half life, but I don't know how to estimate the current activity to find the original. Please help me!!!

    Thank you
     
    Last edited: May 11, 2010
  2. jcsd
  3. May 11, 2010 #2
    The activity is related to the half-life. Activity (A) = [tex]\lambda[/tex] N. [tex]\lambda[/tex] is the decay constant of the element and N is the number of atoms in the sample. Half-life (H) = ln(2)/[tex]\lambda[/tex]. With this, you should be able to rewrite activity in terms of given quantities.
     
  4. May 14, 2010 #3
    Thankyou for that!. . .but I still have one question. . .What's the decay constant? As In how do you calculate it? I think I've been learning by different names. . .
     
  5. May 14, 2010 #4
    The decay constant crops up in exponential decay problems in all sorts of different equations.

    It is in the defining equation of exponential decay [tex]\frac{dN}{dt}=\lambda N[/tex]. The solution to this equation is often written [tex]N=N_0 e^{\lambda t}[/tex].

    You may see it most often in its inverse form as [tex]\tau=1/\lambda[/tex].

    With this relation, it is easy to derive the equation for half-life.

    The decay constant is inversely proportional to the half-life. Half-life=ln(2)/[tex]\lambda[/tex]. If you rearrange that equation, you should be able to find [tex]\lambda[/tex] in terms of a known quantity in this problem, the half-life.

    This is the way to calculate [tex]\lambda[/tex] for this problem, but please do not memorize just that method. There are lots of ways of calculating [tex]\lambda[/tex]. Generally, you will have to look at the equations that you know, find the parameters that you know, and then rearrange those equations to find the parameters that you need.
     
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